How can I round a float value (such as 37.777779) to two decimal places (37.78) in C?
If you just want to round the number for output purposes, then the "%.2f" format string is indeed the correct answer. However, if you actually want to round the floating point value for further computation, something like the following works:
#include <math.h>
float val = 37.777779;
float rounded_down = floorf(val * 100) / 100; /* Result: 37.77 */
float nearest = roundf(val * 100) / 100; /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100; /* Result: 37.78 */
Notice that there are three different rounding rules you might want to choose: round down (ie, truncate after two decimal places), rounded to nearest, and round up. Usually, you want round to nearest.
As several others have pointed out, due to the quirks of floating point representation, these rounded values may not be exactly the "obvious" decimal values, but they will be very very close.
For much (much!) more information on rounding, and especially on tie-breaking rules for rounding to nearest, see the Wikipedia article on Rounding.
Using %.2f in printf. It only print 2 decimal points.
Example:
printf("%.2f", 37.777779);
Output:
37.77
Assuming you're talking about round the value for printing, then Andrew Coleson and AraK's answer are correct:
printf("%.2f", 37.777779);
But note that if you're aiming to round the number to exactly 37.78 for internal use (eg to compare against another value), then this isn't a good idea, due to the way floating point numbers work: you usually don't want to do equality comparisons for floating point, instead use a target value +/- a sigma value. Or encode the number as a string with a known precision, and compare that.
See the link in Greg Hewgill's answer to a related question, which also covers why you shouldn't use floating point for financial calculations.
How about this:
float value = 37.777779;
float rounded = ((int)(value * 100 + .5) / 100.0);
printf("%.2f", 37.777779);
If you want to write to C-string:
char number[24]; // dummy size, you should take care of the size!
sprintf(number, "%.2f", 37.777779);
Always use the printf family of functions for this. Even if you want to get the value as a float, you're best off using snprintf to get the rounded value as a string and then parsing it back with atof:
#include <math.h>
#include <stdio.h>
#include <stddef.h>
#include <stdlib.h>
double dround(double val, int dp) {
int charsNeeded = 1 + snprintf(NULL, 0, "%.*f", dp, val);
char *buffer = malloc(charsNeeded);
snprintf(buffer, charsNeeded, "%.*f", dp, val);
double result = atof(buffer);
free(buffer);
return result;
}
I say this because the approach shown by the currently top-voted answer and several others here -
multiplying by 100, rounding to the nearest integer, and then dividing by 100 again - is flawed in two ways:
For some values, it will round in the wrong direction because the multiplication by 100 changes the decimal digit determining the rounding direction from a 4 to a 5 or vice versa, due to the imprecision of floating point numbers
For some values, multiplying and then dividing by 100 doesn't round-trip, meaning that even if no rounding takes place the end result will be wrong
To illustrate the first kind of error - the rounding direction sometimes being wrong - try running this program:
int main(void) {
// This number is EXACTLY representable as a double
double x = 0.01499999999999999944488848768742172978818416595458984375;
printf("x: %.50f\n", x);
double res1 = dround(x, 2);
double res2 = round(100 * x) / 100;
printf("Rounded with snprintf: %.50f\n", res1);
printf("Rounded with round, then divided: %.50f\n", res2);
}
You'll see this output:
x: 0.01499999999999999944488848768742172978818416595459
Rounded with snprintf: 0.01000000000000000020816681711721685132943093776703
Rounded with round, then divided: 0.02000000000000000041633363423443370265886187553406
Note that the value we started with was less than 0.015, and so the mathematically correct answer when rounding it to 2 decimal places is 0.01. Of course, 0.01 is not exactly representable as a double, but we expect our result to be the double nearest to 0.01. Using snprintf gives us that result, but using round(100 * x) / 100 gives us 0.02, which is wrong. Why? Because 100 * x gives us exactly 1.5 as the result. Multiplying by 100 thus changes the correct direction to round in.
To illustrate the second kind of error - the result sometimes being wrong due to * 100 and / 100 not truly being inverses of each other - we can do a similar exercise with a very big number:
int main(void) {
double x = 8631192423766613.0;
printf("x: %.1f\n", x);
double res1 = dround(x, 2);
double res2 = round(100 * x) / 100;
printf("Rounded with snprintf: %.1f\n", res1);
printf("Rounded with round, then divided: %.1f\n", res2);
}
Our number now doesn't even have a fractional part; it's an integer value, just stored with type double. So the result after rounding it should be the same number we started with, right?
If you run the program above, you'll see:
x: 8631192423766613.0
Rounded with snprintf: 8631192423766613.0
Rounded with round, then divided: 8631192423766612.0
Oops. Our snprintf method returns the right result again, but the multiply-then-round-then-divide approach fails. That's because the mathematically correct value of 8631192423766613.0 * 100, 863119242376661300.0, is not exactly representable as a double; the closest value is 863119242376661248.0. When you divide that back by 100, you get 8631192423766612.0 - a different number to the one you started with.
Hopefully that's a sufficient demonstration that using roundf for rounding to a number of decimal places is broken, and that you should use snprintf instead. If that feels like a horrible hack to you, perhaps you'll be reassured by the knowledge that it's basically what CPython does.
Also, if you're using C++, you can just create a function like this:
string prd(const double x, const int decDigits) {
stringstream ss;
ss << fixed;
ss.precision(decDigits); // set # places after decimal
ss << x;
return ss.str();
}
You can then output any double myDouble with n places after the decimal point with code such as this:
std::cout << prd(myDouble,n);
There isn't a way to round a float to another float because the rounded float may not be representable (a limitation of floating-point numbers). For instance, say you round 37.777779 to 37.78, but the nearest representable number is 37.781.
However, you can "round" a float by using a format string function.
You can still use:
float ceilf(float x); // don't forget #include <math.h> and link with -lm.
example:
float valueToRound = 37.777779;
float roundedValue = ceilf(valueToRound * 100) / 100;
In C++ (or in C with C-style casts), you could create the function:
/* Function to control # of decimal places to be output for x */
double showDecimals(const double& x, const int& numDecimals) {
int y=x;
double z=x-y;
double m=pow(10,numDecimals);
double q=z*m;
double r=round(q);
return static_cast<double>(y)+(1.0/m)*r;
}
Then std::cout << showDecimals(37.777779,2); would produce: 37.78.
Obviously you don't really need to create all 5 variables in that function, but I leave them there so you can see the logic. There are probably simpler solutions, but this works well for me--especially since it allows me to adjust the number of digits after the decimal place as I need.
Use float roundf(float x).
"The round functions round their argument to the nearest integer value in floating-point format, rounding halfway cases away from zero, regardless of the current rounding direction." C11dr ยง7.12.9.5
#include <math.h>
float y = roundf(x * 100.0f) / 100.0f;
Depending on your float implementation, numbers that may appear to be half-way are not. as floating-point is typically base-2 oriented. Further, precisely rounding to the nearest 0.01 on all "half-way" cases is most challenging.
void r100(const char *s) {
float x, y;
sscanf(s, "%f", &x);
y = round(x*100.0)/100.0;
printf("%6s %.12e %.12e\n", s, x, y);
}
int main(void) {
r100("1.115");
r100("1.125");
r100("1.135");
return 0;
}
1.115 1.115000009537e+00 1.120000004768e+00
1.125 1.125000000000e+00 1.129999995232e+00
1.135 1.134999990463e+00 1.139999985695e+00
Although "1.115" is "half-way" between 1.11 and 1.12, when converted to float, the value is 1.115000009537... and is no longer "half-way", but closer to 1.12 and rounds to the closest float of 1.120000004768...
"1.125" is "half-way" between 1.12 and 1.13, when converted to float, the value is exactly 1.125 and is "half-way". It rounds toward 1.13 due to ties to even rule and rounds to the closest float of 1.129999995232...
Although "1.135" is "half-way" between 1.13 and 1.14, when converted to float, the value is 1.134999990463... and is no longer "half-way", but closer to 1.13 and rounds to the closest float of 1.129999995232...
If code used
y = roundf(x*100.0f)/100.0f;
Although "1.135" is "half-way" between 1.13 and 1.14, when converted to float, the value is 1.134999990463... and is no longer "half-way", but closer to 1.13 but incorrectly rounds to float of 1.139999985695... due to the more limited precision of float vs. double. This incorrect value may be viewed as correct, depending on coding goals.
Code definition :
#define roundz(x,d) ((floor(((x)*pow(10,d))+.5))/pow(10,d))
Results :
a = 8.000000
sqrt(a) = r = 2.828427
roundz(r,2) = 2.830000
roundz(r,3) = 2.828000
roundz(r,5) = 2.828430
double f_round(double dval, int n)
{
char l_fmtp[32], l_buf[64];
char *p_str;
sprintf (l_fmtp, "%%.%df", n);
if (dval>=0)
sprintf (l_buf, l_fmtp, dval);
else
sprintf (l_buf, l_fmtp, dval);
return ((double)strtod(l_buf, &p_str));
}
Here n is the number of decimals
example:
double d = 100.23456;
printf("%f", f_round(d, 4));// result: 100.2346
printf("%f", f_round(d, 2));// result: 100.23
I made this macro for rounding float numbers.
Add it in your header / being of file
#define ROUNDF(f, c) (((float)((int)((f) * (c))) / (c)))
Here is an example:
float x = ROUNDF(3.141592, 100)
x equals 3.14 :)
Let me first attempt to justify my reason for adding yet another answer to this question. In an ideal world, rounding is not really a big deal. However, in real systems, you may need to contend with several issues that can result in rounding that may not be what you expect. For example, you may be performing financial calculations where final results are rounded and displayed to users as 2 decimal places; these same values are stored with fixed precision in a database that may include more than 2 decimal places (for various reasons; there is no optimal number of places to keep...depends on specific situations each system must support, e.g. tiny items whose prices are fractions of a penny per unit); and, floating point computations performed on values where the results are plus/minus epsilon. I have been confronting these issues and evolving my own strategy over the years. I won't claim that I have faced every scenario or have the best answer, but below is an example of my approach so far that overcomes these issues:
Suppose 6 decimal places is regarded as sufficient precision for calculations on floats/doubles (an arbitrary decision for the specific application), using the following rounding function/method:
double Round(double x, int p)
{
if (x != 0.0) {
return ((floor((fabs(x)*pow(double(10.0),p))+0.5))/pow(double(10.0),p))*(x/fabs(x));
} else {
return 0.0;
}
}
Rounding to 2 decimal places for presentation of a result can be performed as:
double val;
// ...perform calculations on val
String(Round(Round(Round(val,8),6),2));
For val = 6.825, result is 6.83 as expected.
For val = 6.824999, result is 6.82. Here the assumption is that the calculation resulted in exactly 6.824999 and the 7th decimal place is zero.
For val = 6.8249999, result is 6.83. The 7th decimal place being 9 in this case causes the Round(val,6) function to give the expected result. For this case, there could be any number of trailing 9s.
For val = 6.824999499999, result is 6.83. Rounding to the 8th decimal place as a first step, i.e. Round(val,8), takes care of the one nasty case whereby a calculated floating point result calculates to 6.8249995, but is internally represented as 6.824999499999....
Finally, the example from the question...val = 37.777779 results in 37.78.
This approach could be further generalized as:
double val;
// ...perform calculations on val
String(Round(Round(Round(val,N+2),N),2));
where N is precision to be maintained for all intermediate calculations on floats/doubles. This works on negative values as well. I do not know if this approach is mathematically correct for all possibilities.
...or you can do it the old-fashioned way without any libraries:
float a = 37.777779;
int b = a; // b = 37
float c = a - b; // c = 0.777779
c *= 100; // c = 77.777863
int d = c; // d = 77;
a = b + d / (float)100; // a = 37.770000;
That of course if you want to remove the extra information from the number.
this function takes the number and precision and returns the rounded off number
float roundoff(float num,int precision)
{
int temp=(int )(num*pow(10,precision));
int num1=num*pow(10,precision+1);
temp*=10;
temp+=5;
if(num1>=temp)
num1+=10;
num1/=10;
num1*=10;
num=num1/pow(10,precision+1);
return num;
}
it converts the floating point number into int by left shifting the point and checking for the greater than five condition.
Recently i write a block of code:
const int sections = 10;
for(int t= 0; t < 5; t++){
int i = pow(sections, 5- t -1);
cout << i << endl;
}
And the result is wrong:
9999
1000
99
10
1
If i using just this code:
for(int t = 0; t < 5; t++){
cout << pow(sections,5-t-1) << endl;
}
The problem doesn't occur anymore:
10000
1000
100
10
1
Does anyone give me an explaination? thanks you very much!
Due to the representation of floating point values pow(10.0, 5) could be 9999.9999999 or something like this. When you assign that to an integer that got truncated.
EDIT: In case of cout << pow(10.0, 5); it looks like the output is rounded, but I don't have any supporting document right now confirming that.
EDIT 2: The comment made by BoBTFish and this question confirms that when pow(10.0, 5) is used directly in cout that is getting rounded.
When used with fractional exponents, pow(x,y) is commonly evaluated as exp(log(x)*y); such a formula would mathematically correct if evaluated with infinite precision, but may in practice result in rounding errors. As others have noted, a value of 9999.999999999 when cast to an integer will yield 9999. Some languages and libraries use such a formulation all the time when using an exponentiation operator with a floating-point exponent; others try to identify when the exponent is an integer and use iterated multiplication when appropriate. Looking up documentation for the pow function, it appears that it's supposed to work when x is negative and y has no fractional part (when x is negative and `y is even, the result should be pow(-x,y); when y is odd, the result should be -pow(-x,y). It would seem logical that when y has no fractional part a library which is going to go through the trouble of dealing with a negative x value should use iterated multiplication, but I don't know of any spec dictating that it must.
In any case, if you are trying to raise an integer to a power, it is almost certainly best to use integer maths for the computation or, if the integer to be raised is a constant or will always be small, simply use a lookup table (raising numbers from 0 to 15 by any power that would fit in a 64-bit integer would require only a 4,096-item table).
From Here
Looking at the pow() function: double pow (double base, double exponent); we know the parameters and return value are all double type. But the variable num, i and res are all int type in code above, when tranforming int to double or double to int, it may cause precision loss. For example (maybe not rigorous), the floating point unit (FPU) calculate pow(10, 4)=9999.99999999, then int(9999.9999999)=9999 by type transform in C++.
How to solve it?
Solution1
Change the code:
const int num = 10;
for(int i = 0; i < 5; ++i){
double res = pow(num, i);
cout << res << endl;
}
Solution2
Replace floating point unit (FPU) having higher calculation precision in double type. For example, we use SSE in Windows CPU. In Code::Block 13.12, we can do this steps to reach the goal: Setting -> Compiler setting -> GNU GCC Compile -> Other options, add
-mfpmath=sse -msse3
The picture is as follows:
(source: qiniudn.com)
Whats happens is the pow function returns a double so
when you do this
int i = pow(sections, 5- t -1);
the decimal .99999 cuts of and you get 9999.
while printing directly or comparing it with 10000 is not a problem because it is runded of in a sense.
If the code in your first example is the exact code you're running, then you have a buggy library. Regardless of whether you're picking up std::pow or C's pow which takes doubles, even if the double version is chosen, 10 is exactly representable as a double. As such the exponentiation is exactly representable as a double. No rounding or truncation or anything like that should occur.
With g++ 4.5 I couldn't reproduce your (strange) behavior even using -ffast-math and -O3.
Now what I suspect is happening is that sections is not being assigned the literal 10 directly but instead is being read or computed internally such that its value is something like 9.9999999999999, which when raised to the fourth power generates a number like 9999.9999999. This is then truncated to the integer 9999 which is displayed.
Depending on your needs you may want to round either the source number or the final number prior to assignment into an int. For example: int i = pow(sections, 5- t -1) + 0.5; // Add 0.5 and truncate to round to nearest.
There must be some broken pow function in the global namespace. Then std::pow is "automatically" used instead in your second example because of ADL.
Either that or t is actually a floating-point quantity in your first example, and you're running into rounding errors.
You're assigning the result to an int. That coerces it, truncating the number.
This should work fine:
for(int t= 0; t < 5; t++){
double i = pow(sections, 5- t -1);
cout << i << endl;
}
What happens is that your answer is actually 99.9999 and not exactly 100. This is because pow is double. So, you can fix this by using i = ceil(pow()).
Your code should be:
const int sections = 10;
for(int t= 0; t < 5; t++){
int i = ceil(pow(sections, 5- t -1));
cout << i << endl;
}
Problem description
During my fluid simulation, the physical time is marching as 0, 0.001, 0.002, ..., 4.598, 4.599, 4.6, 4.601, 4.602, .... Now I want to choose time = 0.1, 0.2, ..., 4.5, 4.6, ... from this time series and then do the further analysis. So I wrote the following code to judge if the fractpart hits zero.
But I am so surprised that I found the following two division methods are getting two different results, what should I do?
double param, fractpart, intpart;
double org = 4.6;
double ddd = 0.1;
// This is the correct one I need. I got intpart=46 and fractpart=0
// param = org*(1/ddd);
// This is not what I want. I got intpart=45 and fractpart=1
param = org/ddd;
fractpart = modf(param , &intpart);
Info<< "\n\nfractpart\t=\t"
<< fractpart
<< "\nAnd intpart\t=\t"
<< intpart
<< endl;
Why does it happen in this way?
And if you guys tolerate me a little bit, can I shout loudly: "Could C++ committee do something about this? Because this is confusing." :)
What is the best way to get a correct remainder to avoid the cut-off error effect? Is fmod a better solution? Thanks
Respond to the answer of
David Schwartz
double aTmp = 1;
double bTmp = 2;
double cTmp = 3;
double AAA = bTmp/cTmp;
double BBB = bTmp*(aTmp/cTmp);
Info<< "\n2/3\t=\t"
<< AAA
<< "\n2*(1/3)\t=\t"
<< BBB
<< endl;
And I got both ,
2/3 = 0.666667
2*(1/3) = 0.666667
Floating point values cannot exactly represent every possible number, so your numbers are being approximated. This results in different results when used in calculations.
If you need to compare floating point numbers, you should always use a small epsilon value rather than testing for equality. In your case I would round to the nearest integer (not round down), subtract that from the original value, and compare the abs() of the result against an epsilon.
If the question is, why does the sum differ, the simple answer is that they are different sums. For a longer explanation, here are the actual representations of the numbers involved:
org: 4.5999999999999996 = 0x12666666666666 * 2^-50
ddd: 0.10000000000000001 = 0x1999999999999a * 2^-56
1/ddd: 10 = 0x14000000000000 * 2^-49
org * (1/ddd): 46 = 0x17000000000000 * 2^-47
org / ddd: 45.999999999999993 = 0x16ffffffffffff * 2^-47
You will see that neither input value is exactly represented in a double, each having been rounded up or down to the nearest value. org has been rounded down, because the next bit in the sequence would be 0. ddd has been rounded up, because the next bit in that sequence would be a 1.
Because of this, when mathematical operations are performed the rounding can either cancel, or accumulate, depending on the operation and how the original numbers have been rounded.
In this case, 1/0.1 happens to round neatly back to exactly 10.
Multiplying org by 10 happens to round up.
Dividing org by ddd happens to round down (I say 'happens to', but you're dividing a rounded-down number by a rounded-up number, so it's natural that the result is less).
Different inputs will round differently.
It's only a single bit of error, which can be easily ignored with even a tiny epsilon.
If I understand your question correctly, it's this: Why, with limited-precision arithmetic, is X/Y not the same is X * (1/Y)?
And the reason is simple: Consider, for example, using six digits of decimal precision. While this is not what doubles actually do, the concept is precisely the same.
With six decimal digits, 1/3 is .333333. But 2/3 is .666667. So:
2 / 3 = .666667
2 * (1/3) = 2 * .333333 = .6666666
That's just the nature of fixed-precision math. If you can't tolerate this behavior, don't use limited-precision types.
Hm not really sure what you want to achieve, but if you want get a value and then want to
do some refine in the range of 1/1000, why not use integers instead of floats/doubles?
You would have a divisor, which is 1000, and have values that you iterate over that you need to multiply by your divisor.
So you would get something like
double org = ... // comes from somewhere
int divisor = 1000;
int referenceValue = org * div;
for (size_t step = referenceValue - 10; step < referenceValue + 10; ++step) {
// use (double) step / divisor to feed to your algorithm
}
You can't represent 4.6 precisely: http://www.binaryconvert.com/result_double.html?decimal=052046054
Use rounding before separating integer and fraction parts.
UPDATE
You may wish to use rational class from Boost library: http://www.boost.org/doc/libs/1_52_0/libs/rational/rational.html
CONCERNING YOUR TASK
To find required double take precision into account, for example, to find 4.6 calculate "closeness" to it:
double time;
...
double epsilon = 0.001;
if( abs(time-4.6) <= epsilon ) {
// found!
}
I would have dared say that the numeric values computed by Fortran and C++ would be way more similar. However, from what I am experiencing, it turns out that the calculated numbers start to diverge after too few decimal digits. I have come across this problem during the process of porting some legacy code from the former language to the latter. The original Fortran 77 code...
INTEGER M, ROUND
DOUBLE PRECISION NUMERATOR, DENOMINATOR
M = 2
ROUND = 1
NUMERATOR=5./((M-1+(1.3**M))**1.8)
DENOMINATOR = 0.7714+0.2286*(ROUND**3.82)
WRITE (*, '(F20.15)') NUMERATOR/DENOMINATOR
STOP
... outputs 0.842201471328735, while its C++ equivalent...
int m = 2;
int round = 1;
long double numerator = 5.0 / pow((m-1)+pow(1.3, m), 1.8);
long double denominator = 0.7714 + 0.2286 * pow(round, 3.82);
std::cout << std::setiosflags(std::ios::fixed) << std::setprecision(15)
<< numerator/denominator << std::endl;
exit(1);
... returns 0.842201286195064. That is, the computed values are equal only up to the sixth decimal. Although not particularly a Fortran advocator, I feel inclined to consider its results as the 'correct' ones, given its legitimate reputation of number cruncher. However, I am intrigued about the cause of this difference between the computed values. Does anyone know what the reason for this discrepancy could be?
In Fortran, by default, floating point literals are single precision, whereas in C/C++ they are double precision.
Thus, in your Fortran code, the expression for calculating NUMERATOR is done in single precision; it is only converted to double precision when assigning the final result to the NUMERATOR variable.
And the same thing for the expression calculating the value that is assigned to the DENOMINATOR variable.